Problem 47 - Elementary and Intermediate Algebra

Question

Factor. If a polynomial can't be factored, write "prime." $$ y^{2}-63 $$

Step-by-Step Solution

Answer

The polynomial is prime.

Step 1: Recognize the Structure

The given polynomial is \( y^2 - 63 \). This follows the form \( a^2 - b^2 \), where \( a^2 \) is \( y^2 \) and \( b^2 \) is \( 63 \). Since \( 63 \) is not a perfect square, this can't be immediately factored using simple difference of squares.

Step 2: Check for Simpler Factorization

Try to find perfect squares close to 63. For \( b^2 \) to be a perfect square, \( b \) should be a whole number. Closest perfect squares are 49 (\( 7^2 \)) and 64 (\( 8^2 \)), but neither yield \( 63 \). This suggests that the polynomial is not easily factorable.

Step 3: Conclusion on Factoring

Given that \( y^2 - 63 \) does not fit into a recognizable factoring formula or difference of squares due to 63 not being a perfect square, the polynomial is considered prime over the integers.

Key Concepts

Difference of Squares

The difference of squares is a particular technique in algebra, used specifically to factor expressions structured like \( a^2 - b^2 \). This method is rooted in the identity \( a^2 - b^2 = (a - b)(a + b) \). This powerful algebraic trick allows us to break down certain equations into simpler components, making them easier to solve or analyze.

To effectively apply this, identify whether your expression fits the difference of squares pattern:

• Both terms must be perfect squares: meaning they can be expressed as a square of an integer or variable.
• They must be subtracted from each other.

However, when a term like 63 is involved, complications arise because 63 isn’t a perfect square. This means it cannot be expressed as any integer squared, making it unsuitable for this direct factoring method. So, when pursuing the difference of squares, always check if each term is indeed a perfect square.

Prime Polynomials

Sometimes in algebra, you will encounter polynomials that resist all attempts at factoring—these are referred to as prime polynomials.

Much like prime numbers in arithmetic, which are divisible only by 1 and themselves, prime polynomials can't be broken down into simpler polynomial factors with integer coefficients.

For example, the polynomial \( y^2 - 63 \) is considered prime because:

• The constant term, 63, is not a perfect square or conducive to simpler factoring techniques like the difference of squares.
• No common factor exists that can further decompose the polynomial under integer factorization.

Recognizing a polynomial as prime is crucial since no further algebraic manipulation will break it down with integer factorization.

Integer Factorization

Integer factorization in the context of polynomials refers to the process of breaking down a polynomial into a product of two or more polynomials with integer coefficients. This task is straightforward when dealing with small integers and simple algebraic terms.

However, challenges arise when numbers aren't friendly, like 63 in \( y^2 - 63 \).

• For integer factorization to work, each coefficient or constant in the polynomial should be factorizable into whole numbers.
• If all polynomial components are factored with whole numbers, then integer factorization is successful.
• If one component cannot be factored into a product of integers, the polynomial might be irreducible or prime over the integers.

So, when faced with a situation like \( y^2 - 63 \), where no integer factorization reveals factorable components, the polynomial remains as is, confirming its prime status.